Problem: $\dfrac{ -3f + 6g }{ 2 } = \dfrac{ -5f - 3h }{ -7 }$ Solve for $f$.
Answer: Multiply both sides by the left denominator. $\dfrac{ -3f + 6g }{ {2} } = \dfrac{ -5f - 3h }{ -7 }$ ${2} \cdot \dfrac{ -3f + 6g }{ {2} } = {2} \cdot \dfrac{ -5f - 3h }{ -7 }$ $-3f + 6g = {2} \cdot \dfrac { -5f - 3h }{ -7 }$ Multiply both sides by the right denominator. $-3f + 6g = 2 \cdot \dfrac{ -5f - 3h }{ -{7} }$ $-{7} \cdot \left( -3f + 6g \right) = -{7} \cdot 2 \cdot \dfrac{ -5f - 3h }{ -{7} }$ $-{7} \cdot \left( -3f + 6g \right) = 2 \cdot \left( -5f - 3h \right)$ Distribute both sides $-{7} \cdot \left( -3f + 6g \right) = {2} \cdot \left( -5f - 3h \right)$ ${21}f - {42}g = -{10}f - {6}h$ Combine $f$ terms on the left. ${21f} - 42g = -{10f} - 6h$ ${31f} - 42g = -6h$ Move the $g$ term to the right. $31f - {42g} = -6h$ $31f = -6h + {42g}$ Isolate $f$ by dividing both sides by its coefficient. ${31}f = -6h + 42g$ $f = \dfrac{ -6h + 42g }{ {31} }$